PH 222-2C Fall 2012

Magnetic Fields Due to Currents

Lecture 14

Chapter 29(Halliday/Resnick/Walker, Fundamentals of Physics 8th edition)

1

Chapter 29 Magnetic Fields Due to Currents

In this chapter we will explore the relationship between an electric current and the magnetic field it generates in the space around it. We will follow a two-pronged approach, depending on the symmetry of the problem.

For problems with low symmetry we will use the law of Biot-Savart in combination with the principle of superposition.

For problems with high symmetry we will introduce Ampere’s law.

Both approaches will be used to explore the magnetic field generated by currents in a variety of geometries (straight wire, wire loop, solenoid coil, toroid coil).We will also determine the force between two parallel, current-carrying conductors. We will then use this force to define the SI unit for electric current (the ampere). 2

This law gives the magnetic field generated by a wiresegment of length that carries a current . Considerthe geometry shown in the figure. Associated with theelement

dBds i

ds

The Law of Biot - Savart

we define an associated vector that has magnitude equal to the length . The directionof is the same as that of the current that flows through segment .

dsds

dsds

A

03

The magnetic field generated at point by the element located at point

is given by the equation . Here is the vector that connects 4

point (location of element ) wit

dB P ds Ai ds rdB r

rA ds

7 60

02

h point at which we want to determine .The constant 4 10 T m/A 1.26 10 T m/A and is known as the

sin" " The magnitude of is .4

Here is the angle

P dB

i dsdB dBr

permeability constant.

between and . ds r

034

i ds rdBr

3

0

The magnitude of the magnetic fieldgenerated by the wire at point located at a distance from the wire is given by the equation

.2

iR

P

B

R

Magnetic Field Generated by a LongStraight Wire

0 2

iBR

The magnetic field lines form circles that have their centers

at the wire. The magnetic field vector is tangent to the

magnetic field lines. The sense for is given by the. We p

B

B right -hand rule

oint the thumb of the right hand in the direction of the current. The direction along which the fingers of the right hand curl around the wire gives the

direction of .B

4

02

Proof of the equation. Consider the wire element of length shown in the figure.The element generates at point a magnetic field of

sinmagnitude . Vector is pointing 4

the p

dsP

i dsdB dBr

into

02

0 0

2 2 2 2

age. The magnetic field generated by the whole wire is found by integration:

sin22

sin sin / /

i dsB dB dBr

r s R R r R s R

3/2

0 0 03/2 2 2

2 2 22 2

2 20 0

2 2 2i i

dx xa x a

iRd

x

s sBR Rs Rs R

a

0 2

iBR

5

. .dB

B Magnetic Field Generated by

a Circular Wire Arc of Radius at Its Center R C

0 02 2

0

A wire section of length generates at the center a magnetic field .sin 90The magnitude . The length

4 4

. Vector points of the page.4

The

ds C dBi ids dsdB ds Rd

R RidB d dBR

out

0 0

0ci

0

rc

net magnetic field =4

The angle must be expressed in radians.

For a circular wire, 2 . In this case we get:

.4

.2

iB d iB d

i

R R

BR

Note :

0 4

iBR

6

0 02 2

The wire element generates a magnetic field sin 90whose magnitude .

4 4We decompose into two components:One

ds dBi ids dsdB

r rdB

Magnetic Field Generated by a Circular Loop Along the Loop Axis

( ) is along the -axis.The second component ( )is in a direction perpendicular to the -axis. The sum of all the is equal to zero. Thus we sum only the terms:

dB z dBz

dB dB

2

03/22

2 2 02

0 03/23 2 2

0 03/2 3/22 22 2 2

cos cos

2

cos4

4 4

24 4

i ds Rr R z dB dBr r

iR iRds dsdB B dBr R z

iR iRB ds RR z R z

iR

R z

7

8

Example

µo- permiability constant

1221 BLIF

9

Superposition Principle

10

The law of Biot-Savart combined with theprinciple of superposition can be used to

determine if we know the distributionof currents. In situations that have highsymmetry we can use Amper

B

Ampere's Law

e's law instead, because it is simpler to apply.

Ampere's law can be derived from the law of Biot-Savart, with which it ismathematically equivalent. Ampere's law is more suitable for advanced formulations of electromagnetism. It can be expressed as follows:

0

The line integral of the magnetic field along any closed path

is equal to the total current enclosed inside the path multiplied by .

B ds B

The closed path used is known as an " " In its present formAmpere's law is not complete. A missing term was added by Clark Maxwell.The complete form of Ampere's law will be discussed in

Amperian loop.

Chapter 32.

0 encB ds i

11

0 encB ds i

1 2

1 1

Determination of . The closed path is

divided into elements , , ..., .We then form the sum:

cos . Here is the

ma

n

n n

i i i i i ii i

B ds

n s s s

B s B s B

Implementation of Ampere's Law :

1.

1

enc

gnetic field in the th element.

lim as

Calculation of . We curl the fingersof the right hand in the direction in which the Amperian loop was traversed. We notethe

n

i ii

i

B ds B s n

i

2.

direction of the thumb.

All currents inside the loop to the thumb are counted as .All currents inside the loop to the thumb are counted as .All currents outside the loop are not count

paantiparal

rallel positivelel negative

enc 1 2

ed. In this example : .i i i 12

We already have seen that the magnetic field linesof the magnetic field generated by a long straight wire that carries a current have the form of circles,

i

Magnetic Field Outside a Long Straight Wire

which are concentric with the wire.We choose an Amperian loop that reflects thecylindrical symmetry of the problem. Theloop is also a circle of radius that has its centeron the wire. The magnetic

r

0 enc

0

0

field is tangent to the loop and has a constant magnitude :

cos 0 2

Ampere's law holds true for any closed path.We choose to use the path that makes h

2

t e

B

B ds Bds B ds rB i

iBr

i

Note :

calculation

of as easy as possible.

B

13

We assume that the distribution of the currentwithin the cross-section of the wire is uniform.The wire carries a current and has radius .We choose an Amper

i R

Magnetic Field Inside a Long Straight Wire

0 enc

2 2

enc 2 2

ian loop that is a circle of radius ( ) with its centeron the wire. The magnetic field is tangent to theloop and has a constant magnitude :

cos 0 2

r r R

B

B ds Bds B ds rB i

r ri i iR R

2

0 20

222 rrB iB r

RiR

R

0

2iR

r

B

O 14

The solenoid is a long, tightly wound helical wire coil in which the coil length is much larger than thecoil diameter. Viewing the solenoid as a collectionof single circular loops, one

The Solenoid

can see that the magneticfield inside is approximately uniform.

The magnetic field inside the solenoid is parallel to the solenoid axis. The sense of can be determined using the right-hand rule. We curl the fingers of the right handalong the direction of the c

B

urrent in the coil windings. The thumb of the right hand

points along . The magnetic field outside the solenoid is much weaker and can be taken to be approximately zero.

B

15

We will use Ampere's law to determinethe magnetic field inside a solenoid. Weassume that the magnetic field is uniforminside the solenoid and zero outside. Weassume that the solenoid has turns perun

nit length.

We will use the Amperian loop . It is a rectangle with its long side parallelto the solenoid axis. One long side ( ) is inside the solenoid, while the other ( )

is outside: b

a

abcdab cd

B ds B ds

enc

0 e c 0n 0

cos 0 0

The enclosed current .

c d a

b c db b b c d a

a a a b c d

B ds B ds B ds

B ds Bds B ds Bh B ds B ds B ds

B ds Bh i nhi

B ds i Bh nhi B ni

0B ni

16

A toroid has the shape of a doughnut (see figure).We assume that the toroid carries a current and thatit has windings. The magnetic field lines inside the toroidform cir

iN

Magnetic Field of a Toroid

cles that are concentric with the toroid center.The magnetic field vector is tangent to these lines.

The sense of can be found using the right-hand rule.We curl the fingers of the right handalong th

B

e direction of the current in the coil windings.

The thumb of the right hand points along . The magneticfield outside the solenoid is approximately zero.

B

enc

00

We use an Amperian loop that is a circle of radius (orange circle in the figure):

cos 0 2 . The enclosed current .

Thus: 2

T

.

he magnetic f2

iel

r

B ds Bds B ds rB i Ni

NiBr

rB Ni

Note :

d inside a toroid is not uniform.

2oNiB

r

17

20

2

Consider the magnetic field generated by a wire coil of radius that carries a current . The magnetic fieldat a point on the -axis is given by

2

R iP ziRB

R z

The Magnetic Field of a Magnetic Dipole.

3/ 22

20

3

20 0 0

3 3 3

. Here is the distance between

and the coil center. For points far from the loop

( ) we can use the approximation: 2

. Here is the magnetic2 2 2

dipole

z

PiRz R Bz

i R iABz z z

03

moment of the loop. In vector form:

( ) .2

The loop generates a magnetic field that has the sameform as the field generated by a bar magnet.

B zz

03( )

2B z

z

18